Let us consider Q2 which can have only two correct answers, A or D. If its A then it forces the following answers Q: 1 2 3 4 5 6 7 8 ; A: C A C D D D and now none of the four answers for Q3 will be consistent with the others. If the answer for Q2 is D then Q: 1 2 3 4 5 6 7 8 A: D C D This leaves us with B as the only choice for Q1, Q: 1 2 3 4 5 6 7 8 A: B D B C A D But this cannot give a score of 8 since the answer for Q6 is wrong. Thus a score of 8 is not possible.

Now, let us suppose that there is a solution with 7 correct. Either 8B is the right option or all the options inclusive of 8 are incorrect. Let's try the latter path. Since it implies 1-7 are correct, all of the logic in the 8-correct analysis still holds, up to the point where 3D implies 8A. We have: 1C 2A 3D 4C 7B 8A are correct while5B and 6B . By 4C we need two more Ds, so 5D and 6D. This turns out to be a consistent solution with Q8 being the only wrong answer.

Consequently, 1C-2A-3D-4C-5D-6D-7B-8A is the combination with the unique highest possible score of 7, and this has been achieved by answering the last question wrong.